Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T =
Temperature = 120.21°C
c). Specific volume
d). Specific energy ()
e). Specific enthalpy
At
f). Enthalpy at m = 0.5 kg
= 1022.91 kJ
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂